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(D) Consider a small differential element $dl$ on the semi-circular ring. The charge $dq$ on this element is given by $dq = \lambda dl$,where $\lambda

Vedclass · Accepted Answer

$-\frac{q}{2\pi^2\varepsilon_0 r^2}\hat{j}$

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(D) Consider a small differential element $dl$ on the semi-circular ring. The charge $dq$ on this element is given by $dq = \lambda dl$,where $\lambda = \frac{q}{\pi r}$ is the linear charge density.Since $dl = r d	heta$,we have $dq = \left(\frac{q}{\pi r}ight) (r d	heta) = \frac{q}{\pi} d	heta$.The magnitude of the electric field $dE$ at the centre $O$ due to this element is $dE = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q}{\pi r^2} d	heta = \frac{q}{4\pi^2\varepsilon_0 r^2} d	heta$.Due to symmetry,the horizontal components $(dE \cos	heta)$ of the electric field from elements on opposite sides of the vertical axis cancel each other out.The net electric field is the integral of the vertical components $(dE \sin	heta)$ from $	heta = 0$ to $\pi$.$E = \int_0^{\pi} dE \sin	heta = \int_0^{\pi} \frac{q}{4\pi^2\varepsilon_0 r^2} \sin	heta d	heta$.$E = \frac{q}{4\pi^2\varepsilon_0 r^2} [-\cos	heta]_0^{\pi} = \frac{q}{4\pi^2\varepsilon_0 r^2} (-(-1) - (-1)) = \frac{q}{4\pi^2\varepsilon_0 r^2} (2) = \frac{q}{2\pi^2\varepsilon_0 r^2}$.Since the charge is positive and distributed on the upper semi-circle,the field at the centre points downwards,i.e.,in the $-\hat{j}$ direction.Therefore,$\vec{E} = -\frac{q}{2\pi^2\varepsilon_0 r^2} \hat{j}$.

$A$ thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net electric field $\vec{E}$ at the centre $O$ is

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